在 Lua 4 中将表值作为函数参数传递，无需“调用"功能

问题描述

我有一个原则上可以是任意长度的值表:

I have a table of values that can in principle be of any length:

Points =
{
    "Point #1",
    "Point #5",
    "Point #7",
    "Point #10",
    "Point #5",
    "Point #11",
    "Point #5",
}

我想将它们作为参数传递给函数.

I want to pass them as arguments to a function.

addPath(<sPathName>, <sPoint>, <sPoint>, ...)

现在，通常您可以使用呼叫"功能.但在我使用的软件中，此功能不可用且不在范围内.

Now, normally you could use the "call" function. But in the software I am using this function is unavailable and not in scope.

如何在 Lua 4 中解决这个问题?

How do I get around this problem in Lua 4?

这里是我可以使用的功能.

推荐答案

在较新版本的 Lua 中，您将使用 unpack，如 addPath(sPathName,unpack(Points))，但 Lua 4.0 没有 unpack.

In newer versions of Lua, you'd use unpack, as in addPath(sPathName,unpack(Points)), but Lua 4.0 does not have unpack.

如果您可以添加 C 代码，Lua 5.0 中的 unpack 在 4.0 中可以正常工作:

If you can add C code, unpack from Lua 5.0 works fine in 4.0:

static int luaB_unpack (lua_State *L) {
  int n, i;
  luaL_checktype(L, 1, LUA_TTABLE);
  n = lua_getn(L, 1);
  luaL_checkstack(L, n, "table too big to unpack");
  for (i=1; i<=n; i++)  /* push arg[1...n] */
    lua_rawgeti(L, 1, i);
  return n;
}

将此添加到 lbaselib.c 并将其添加到 base_funcs:

Add this to lbaselib.c and this to base_funcs:

  {"unpack", luaB_unpack},

如果您不能添加 C 代码，那么您就不走运了，可能会沦为这种黑客:

If you cannot add C code, then you're out of luck and are probably reduced to this hack:

function unpack(t)
  return t[1],t[2],t[3],t[4],t[5],t[6],t[7],t[8],t[9],t[10]
end

根据需要扩展返回表达式，但最多只能扩展到 200 左右.让我们希望 addPath 忽略或停在第一个 nil.

Extend the return expression as needed but you may only go as far as 200 or so. Let's hope that addPath ignores or stops at the first nil.

你也可以试试这个，它在第一个 nil 处停止，但没有明确的限制(有递归限制，它最多只能处理 250 个表条目):

You can also try this one, which stops at the first nil but has no explicit limits (there are recursion limits and it'll only handle up to 250 table entries):

function unpack(t,i)
        i = i or 1
        if t[i]~=nil then
                return t[i],unpack(t,i+1)
        end
end

这篇关于在 Lua 4 中将表值作为函数参数传递，无需“调用"功能的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！